Calculate the size of fuses to use

[yungman]

TL;DR Summary

Calculate the size of the fuses to use

Hi

This is what Chinese saying "Old cat got his whiskers burn"! All these years, I never even have to worry power consumption to put in fuses. This is for the power amp I designed and I want to put in fuses for protection.

The peak output voltage is 36V driving a 3ohm load. Output is a sine wave. I have F1 as AC line input fuse and F2 & F3 are fuses for the +ve and -ve supply rail of the amp. Here is my calculation. I want to run it by you experts to make sure I am correct:

Please let me know am I right.

Thanks

Alan

 

[yungman]

Anyone? I just want to confirm, I am 90% sure, but just want to confirm.

thanks

 

[berkeman]

Ack, please, please, please learn to post in LaTeX. It is very hard to decipher your work in pictures. I know you are very new here, and may not be familiar with this PF rule about posting in LaTeX. There is a "LaTeX Guide" link at the lower left of the Edit window.

Also, as I try to decipher your images, it looks like you calculate the load current as 12A max, but want to use a 3A-7A fuse?

 

[yungman]

Ack, please, please, please learn to post in LaTeX. It is very hard to decipher your work in pictures. I know you are very new here, and may not be familiar with this PF rule about posting in LaTeX. There is a "LaTeX Guide" link at the lower left of the Edit window.

Also, as I try to decipher your images, it looks like you calculate the load current as 12A max, but want to use a 3A-7A fuse?

I forgot how to use latex! 12A is peak current, also it's only conduct 1/2 of the cycle. The negative half draws current from the negative supply that has it's own fuse(F3). So the peak current is only equal to 6A. then multiply by 0.707 to get 4.24A.

In another word, the total power is split between the +ve and -ve supply, each only has to provide half of the total power.

Thanks

 

[DaveE]

Sorry, I haven't read your post in detail. But, a general comment that few outside of power electronics understand about fuses.

According to safety standards and common practice they are there to prevent fire or (maybe) insulation breakdown due to high temperature due to excess current. They are to protect against excess current in fault conditions and actually have little to do with normal circuit operation. How much current your circuit draws is only of peripheral relevance. The details are a bit too much to explain here but the normal (most common) design process goes like this:

1) My circuit draws x amps, and I need wires (and such) to reliably supply x amps without getting too hot or dropping too much voltage.

2) The wires I've chosen will overheat (above the temp. rating of the insulation) if they experience y amps of current.

3) I need to select a fuse (or other device) that can reliably pass x amps and will disconnect below y amps.

You choose fuses based on the wire size*, not your circuit draw. The wire size is selected based on the circuit(s) it needs to power. Wire size is selected to make stuff work normally. Fuses are selected to protect the wire insulation. Every EE that does this has their favorite table to help. But the safety inspectors don't care about your tables, they'll want to measure the temperature of the insulation in normal operation worst case conditions. Then the will create a reasonable high current fault, and see if you've protected it from bad consequences. They may just skip the first test, but not the second one.

My laptop charger might be drawing about 1 amp right now. But it's plugged into a circuit that has wire to support >20 amps and a circuit breaker that will allow 20 amps, but not so much more that the wire insulation, outlets, etc. will be damaged.

One of the most common, and not so easy, questions in equipment design is what wire size and fuse should I use. It is actually often a quite complex thermal problem. You can spot a pro if their first question is "what is the temperature rating of the wire insulation, and what's nearby that also makes heat".

* OK, sometimes fuses are (were?) selected to protect equipment, e.g. "rectifier fuses". But that's not common, and often, IMO, a fix for a bad or difficult design.

 

[yungman]

Thanks

I since decided to use 8A fuse for F2 and F3. The output stage has over 9 pairs of power transistor that easily supply 20A+. 8A fuse should cover all normal operation and will never blow under normal condition.

All my wires are 12 gauge, wire conductor is not going to be a problem. Speaker connection is 10gauge, it definitely will not over heat with 8A.

As for AC input fuse F1. From calculation, max power draw from each channel is 216W, so both channels is about 440W max. So say I have to be able to sustain 500W.

P=Irms X Vrms => Irms= P/Vrms = 500W/110V = 4.54545A. So a 5A fuse should do it.

 

[Tom.G]

As for AC input fuse F1. From calculation, max power draw from each channel is 216W, so both channels is about 440W max. So say I have to be able to sustain 500W.

P=Irms X Vrms => Irms= P/Vrms = 500W/110V = 4.54545A. So a 5A fuse should do it.

Don't forget inrush surge current.

Little details -- like charging the filter caps, or switching the power on at peak voltage when the core of the power transformer is already magnetized in the same direction -- can cause nuisance blowing without a slow-blo fuse.

Cheers,
Tom

 

[yungman]

Don't forget inrush surge current.

Little details -- like charging the filter caps, or switching the power on at peak voltage when the core of the power transformer is already magnetized in the same direction -- can cause nuisance blowing without a slow-blo fuse.

Cheers,
Tom

Thanks

I use slow-blo fuse and I have 2.5ohm 8A inrush current limiter for each transformer also. I did that more to protect the contacts of the power switch and I also put a 0.1uF in series with 10ohm resistor across the two terminals of the power switch to snub out the inductor spike from the transformer.

 

[Tom.G]

I did that more to protect the contacts of the power switch and I also put a 0.1uF in series with 10ohm resistor across the two terminals of the power switch to snub out the inductor spike from the transformer.

Those are almost the "classic values"; 100 Ohm is more common -- though dependent on switch rating and load current and inductance. The 10 Ohm could allow 17 Amps thru the switch contacts if closed at the voltage peak.

Cheers,
Tom

 

[yungman]

Those are almost the "classic values"; 100 Ohm is more common -- though dependent on switch rating and load current and inductance. The 10 Ohm could allow 17 Amps thru the switch contacts if closed at the voltage peak.

Cheers,
Tom

Are you talking about the snubber network of 10ohm in series with 0.1uF across the two terminals of the power switch? I thought that's only for inductive spike from the inductance of the transformers when the switch is opened.

Are you talking about when closing at the peak voltage (117Vrms X 1.414 = 165Vp) where the 0.1uF is charged to 165V and short out by the switch through the 10ohm?

I really don't know how to calculate exactly all the current. I know the thermistor limit the current, the inductance of the transformer also limit the in rush current. But on the secondary of the transformer, I have about 0.2F(yes 0.2 farads) of filter caps. I really don't even know how to start the calculation. All I can say is I have amps like this and it's been 4 years and never burn one switch yet! I know, it's not very scientific!

 

[Tom.G]

4 years and never burn one switch yet! I know, it's not very scientific!

In that case, I say the devil with "scientific" as you are obviously doing other things (that you have not mentioned) right!

Keep up the good work.

Cheers,
Tom